208k views
1 vote
A circuit consists of a battery, a 100 k2 resistor, a 20.0 uF capacitor in series with a switch which is initially in the open position. The capacitor is initially uncharged

If the EMF of the battery is ε= 3 3 . 6 v ε=33.6 v, calculate the current through the resistor 6.00 seconds after the switch is closed.

User Igor Minar
by
8.4k points

1 Answer

4 votes

Final answer:

The current through the resistor 6.00 seconds after the switch is closed in a series circuit with a resistor, uncharged capacitor, and EMF, is found using the formula for the charging of a capacitor in an RC circuit with known values for resistance, capacitance, EMF, and time.

Step-by-step explanation:

The student's question involves a circuit with a resistor, a capacitor, and a battery. The student is asked to calculate the current through the resistor 6.00 seconds after the switch is closed, in a case where the capacitor is initially uncharged and an EMF of 33.6 V is provided by the battery. To find the current, we can use the formula for the charging of a capacitor through a resistor in an RC circuit:

I(t) = \( \frac{\epsilon}{R} \cdot e^{-\frac{t}{RC}} \)

Where:

  • \(I(t)\) is the current at time t,
  • \(\epsilon\) is the EMF of the battery,
  • R is the resistance,
  • C is the capacitance, and
  • t is the time after the switch is closed.

Plugging in the given values:

R = 100 k\Omega = 100,000 \Omega

C = 20.0 \mu F = 20.0 \cdot 10^{-6} F

\(\epsilon\) = 33.6 V

t = 6.00 s

The time constant is \(RC = 100,000 \Omega \cdot 20.0 \cdot 10^{-6} F = 2\) s, so:

I(6.00 s) = \( \frac{33.6}{100,000} \cdot e^{-\frac{6.00}{2}} \)

From here, the student can calculate the current at time t = 6.00 seconds using the exponential decay function to find the final answer.

User Zrajm
by
7.8k points